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hi seek This is a basic structure of any multiplication principle : I will put it down ; if an event can occur in " m " diffeent ways and if following it a second event can occur in " n "diferent ways , then two events in succession can occur in m X n ways ....

e.g in a cinema there are 3 entrance door and 2 exit doors ..then a person can have 3 * 2 = 6 possible routes ....

I have taken this explantion from the introduction of Permutation topic . If you want more help I can scan a few pages from a book which has about 10 -15 question and answers on this topic ...

similar question can be how many 3 digit odd numbers can be formed by using the digits 1 2 3 4 5 6 such that there is no repetition ....

I have been trying to solve the question with the digits not being repeated. This is how it goes: Case 1: One Digit Number: 4 ways Case 2: Two Digits Numbers: 4*3 = 12 ways Case 3: Three Digits Numbers: 3*3*2 = 18 ways Therefore, total number of ways = 4+12+18= 34. Am I correct? I think I am and therefore I was wondering how the answer could be 60 when the digits are not repeated.

In case the digits are repeated: Case 1: One Digit Numbers: 4 ways Case 2: Two Digits Numbers: 4*4 = 16 ways Case 3: Three Digits Numbers: 3*4*4 = 48 ways Therefore, total number of ways = 4+16+48 = 68.

Last edited by shekharvineet on 24 Aug 2010, 09:55, edited 1 time in total.

Hi. Here it goes. In case the digits are repeated: Case 1: One Digit Numbers: It can be filled by any of those four numbers in 4 different ways. Case 2: Two Digits Numbers: The tens' place can be filled by those four numbers in 4 different ways. Similarly the units' place can be filled in 4 different ways, since the numbers can be repeated. So, number of ways of filling two digits numbers = 4*4 = 16. Case 3: Now there is a restriction here. We are to find positive integers less that 500, so 5 caanot be used to fill up the hundreds' place. It can be filled in only 3 different ways. But the tens' place and units' place each can be filled in 4 different ways. So, number of ways of filling three digitd numbers= 3*4*4 = 48. Hence, total number of ways = 4+16+48 = 68.

_ 4 ways _ _ 4*3 ways _ _ _ (1st digit cant take 5 as it should be less than 500 so it 3 ways )3*4*3 4+12+36=52

if it is repeated than it is 68 ways

_ 4 ways

_ _ 4*4 ways =16

_ _ _ 3*4*4=48

48+16+4=68

I don't think your answer is correct when the digits are not repeated. That is , the correct answer is 34 and not 52.

Now we agree that in case of One digit numbers, it can be filled in 4 ways; and in case of two digit numbers, it can be filled in 4*3= 12 ways. But I beg to differ when it comes to three digit numbers. I say that 5 cannot be used to fill the hundreds place because there is a restriction that the number has to less than 500. How can you even allow 5 to fill the hundreds place because the question clearly says it has be less than 500, i.e till 499. Even if we change the question and ask to find the number of ways the digits 1, 2, 3 and 5 can be used to form numbers less than or equal to 500, and allow 5 to occupy the hundreds place, how will you fill up the remaining two places, i.e. the tens place and the units place because the other three digits are 1, 2 and 3 which are all greater than 0. So if you allow 5 to fill the hundreds place, it voilates the entire question. So, there are only 3*3*2 = 18 ways of filling in case of three digit numbers.

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In this case to include 500 we need zeroes as possible digits but we do not have them, so there is no difference between <500 and <501. Is that right?

Yes, that's correct.

Hi Bunuel,

I could not get the logic behind this. Can you please explain? Is there any other easy way to solve such questions?

How many positive integers less than 500 can be formed using the numbers 1, 2, 3 and 5 for the digits?

(A) 48 (B) 52 (C) 66 (D) 68 (E) 84

Positive integers less than 500 can be:

1. A single-digit integer: 4 2. A double-digit integer: 4*4. 3. A three-digit integer: 3*4*4 (the hundreds digit cannot be 5, so we have only 3 options for it).

Re: How many positive integers less than 500 can be formed using [#permalink]

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25 Apr 2016, 21:43

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How many positive integers less than 500 can be formed using the numbers 1, 2, 3 and 5 for the digits?

(A) 48 (B) 52 (C) 66 (D) 68 (E) 84

HI,

you could do these Qs in three ways--

1) find out separately for 2-, 3- and 1-digit number.. a) 1-digit - 1,2,3,5 - 4 ways b) 2-digit - 4*4 = 16 ways c) 3-digit - 3*4*4 = 48.. we cannot have 5 in hundreds place so 3 instead of 4.. total = 4+16+48 = 68..

2) say we use 0 in first place, it will include BOTH 2 and 3- digits.. a) single digit - 1,2,3,5 - 4 ways.. b) 2- and 3-digits = 4*4*4 = 64 .. here hundreads can be by 0,1,2,3 and tens and unit can be by 1,2,3,5 Total = 4+64 =68

3) we consider 0 in both hundreds and tens place, it caters for all 1-,2- and 3-digit numbers BUT adds 0 in tens place in three digit numbers.. a) all = 4*5*4= 80.. b) subtract 0 in tens place in 3-digits = 3*1*4= 12.. total = 80-12=68
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